Base | Representation |
---|---|
bin | 101101010010011101110… |
… | …100001011000101100001 |
3 | 102000112010211221011202211 |
4 | 231102131310023011201 |
5 | 401442243212201423 |
6 | 10341420553042121 |
7 | 440564213465014 |
oct | 55223564130541 |
9 | 12015124834684 |
10 | 3112204022113 |
11 | a9a97304326a |
12 | 4231ba9b6341 |
13 | 1976308390b4 |
14 | aa8baac337b |
15 | 55e500a010d |
hex | 2d49dd0b161 |
3112204022113 has 8 divisors (see below), whose sum is σ = 3149046874496. Its totient is φ = 3075574752000.
The previous prime is 3112204022077. The next prime is 3112204022141.
3112204022113 is nontrivially palindromic in base 10.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 3112204022113 - 237 = 2974765068641 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (3112204022153) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 53366253 + ... + 53424538.
It is an arithmetic number, because the mean of its divisors is an integer number (393630859312).
Almost surely, 23112204022113 is an apocalyptic number.
It is an amenable number.
3112204022113 is a deficient number, since it is larger than the sum of its proper divisors (36842852383).
3112204022113 is a wasteful number, since it uses less digits than its factorization.
3112204022113 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 106791135.
The product of its (nonzero) digits is 576, while the sum is 22.
Multiplying 3112204022113 by its sum of digits (22), we get a palindrome (68468488486486).
The spelling of 3112204022113 in words is "three trillion, one hundred twelve billion, two hundred four million, twenty-two thousand, one hundred thirteen".
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